If $f(x)=\sqrt x-\sqrt {4-x}+2$, what is the domain of $\sqrt{f(x)-f^{-1}(x)}$?

Question

If $f(x)=\sqrt x-\sqrt {4-x}+2$, what is the domain of $\sqrt{f(x)-f^{-1}(x)}$?

We have $f(x)\ge f^{-1}(x)$. Since the graph of $f^{-1}(x)$ and $f(x)$ are symmetric along the line $y=x$, intuitively I realized that in order to $f(x)\ge f^{-1}(x)$ we should have $f(x)\ge x$,

$$\sqrt x-\sqrt{4-x}+2\ge x \Rightarrow \sqrt x-\sqrt{4-x} \ge x-2$$Here I'm not sure how to continue. Squaring doesn't help a lot (since it is valid only for $x\ge2$ and after simplification and squaring again one would get a fourth degree polynomial). I noticed both functions $g(x)=\sqrt x-\sqrt {4-x}$ and $h(x)=x-2$ are strictly increasing but I don't know how to use this fact.

Answer 1

Hint

The plot of $f(x)$ and $g(x)=x$ gives you a very good idea about how to proceed. Your idea about looking for $f(x)\ge x$ is good, looking at the plotting.

Also see that even if you have a fourth degree polynomial, you can guess at lest two roots. I would try $0, 2, 4$ as roots.

Can you finish?